JEE Main Physics question
A spring of force constant 800 N m⁻¹ has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is
Answer: (B)
Solution
(Same as Q27.) W = ½k(x₂²−x₁²) = ½(800)(0.15²−0.05²) = 8 J.
A spring of force constant 800 N m⁻¹ has an extension of 5 cm. The work done in extending it from 5 cm to 15 cm is
(Same as Q27.) W = ½k(x₂²−x₁²) = ½(800)(0.15²−0.05²) = 8 J.